Linear depedency

If not, then there must exist two scalars, a and b (they can't both be zero), such that

[tex]a e^x + b e^{2x} = 0[/tex]

for all x. Can you find such scalars?

 

Since that is to be true for all x, you can get two simple equations to solve for a and b by choosing two values of x.

 

I think it's possible:
[tex]a e^x + b e^{2x} = 0[/tex] <==> [tex]a e^x = - b e^{2x}[/tex] <==> [tex]ln (a e^x) = - ln (b e^{2x})[/tex] <==> ax = -2bx

So if we carefully choose a, we can definitely show your given set is linearly dependent.

 

Except

[tex]\ln(a e^x) = \ln(a) + x[/tex]
[tex]\ln(-b e^{2x}) = \ln(-b) + 2x[/tex]

so you need

[tex]\ln(a) - \ln(-b) = x[/tex] for all x!

 

Bummer, so sorry.

 

No, not correct. You need to show that a=b=0. That is the DEFINITION of linear dependence.

 

Apologies, I missed out an 'in'. The functions e^x and e^2x are clearly linearly independent (over R), which is where I misunderstood what you're saying. (If you don't see it then just rearrange ae^x +be^2x=0 to see that you're claim of linear dependence implies that e^x = -a/b for all x.)

 

No, that is incorrect. You have shown that taking a= -b makes [itex]ae^x+ be^{2x}= 0[/itex] for x= 0. To show they are linearly dependent, you would have to find a and b so that was true for ALL x. And you can't. Taking x= 0 gives a+ b= 0, so a=-b, while taking x= 1 gives ae+ be^2= 0 so a= -be. Those two equations are only possible if a= b= 0.

Another way to prove it is this: if [itex]ae^x+ be^{2x}= 0[/itex] for all x, then it is a constant and its derivative must be 0 for all x: [itex]ae^x+ 2be^{2x}= 0[/itex]. Taking x= 0 in both os those, a+ b= 0 and a+ 2b= 0. Again, those two equations give a= b= 0.

 

Yes, that is why you can't "prove [itex]e^x[/itex] and [itex]e^{2x}[/itex] are dependent": they are independent.

 

Remember this is equal to the zero *function* in the vector space of real valued functions. The zero function is the one that is zero for all its inputs.

 

That isn't right. You have your quantifiers all kludged up into one it should have read

(for all a,b)(ae^x + be^2x=0 for all x => a=b=0)

the negation of which is

(there exists a,b)(ae^x + be^2x=0 for all x AND NOT(a=b=0))

The for all x is not part of the definition of linear (in)dependence, it is part of the definition of what it means for the function to be zero.

 

In what way is that not in a useful quantifier form (for those who like things like that)? Really, we shouldn't even bother with the quantifier "for all a,b", and putting things in the full on formal abstract quantifier notation just makes things far more opaque than they need to be.

You're making a very easy question seem very hard: can you find real numbers a and b not both equal to 0 so that

ae^x + be^2x

is the zero function? No - it has been proven several times in this thread.

 

The point is that you are working in a vector space of functions. When we say "af(x)+ bg(x)= 0" we mean it is equal to the 0 function- 0 for all x.